What is the horizontal range of a projectile fired into the air?

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To determine the horizontal range of a projectile fired into the air, the velocity vector at 3.3 seconds is given as v = (8.6i + 5.0j) m/s. The horizontal component of the velocity (Vx) is 8.6 m/s, and the time of flight is 3.3 seconds. The horizontal range can be calculated using the formula R = Vx * t, resulting in R = 8.6 * 3.3, which equals 28.38 meters. The initial attempt to calculate the range using the resultant velocity was incorrect, as the horizontal range specifically depends on the horizontal component of the velocity. Understanding the distinction between horizontal and vertical components is crucial for solving projectile motion problems accurately.
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Homework Statement


Exactly 3.3s after a projectile is fired into the air from the ground, it is observed to have a velocity v = (8.6i + 5.0j ), where the x-axis is horizontal and the y-axis is positive upward.

Determine the horizontal range of the projectile

Homework Equations


x=x0+vy0t+1/2at2
v=\sqrt{}V<sub>x</sub>2+Vy2

The Attempt at a Solution


v= 9.95 by finding the resultant of the velocity vector
x 0 = 0
ax= 0
x= Vx(t)?
is that correct?
if so then,
x= 9.95(3.3)= 32.8?
but this is wrong and i don't know how to get it
wait does R= x? or y?
 
Last edited:
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Use the information that v = (8.6i+5.0j)m/s in 3.3 s to find the components of the initial velocity vector. Then use the equation that relates the range R to these initial components to answer the question.
 

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